Asked by rei

An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 9.6 m lower vertically, is a horizontally situated spring with constant 4.7 × 10^
5 N/m.
The acceleration of gravity is 9.8 m/s^
2. Ignore friction.
How much is the spring compressed in stopping the ore car? Answer in units of m.
Answered by drwls
Gravitational potential energy loss = Spring potential energy increase

M g H = (1/2) k X^2

k = 4.7*10^5 N/m
M = 4.1*10^4 kg
H = 9.6 m
You known what g is.

Solve for X
Answered by rei
Thank you!
Answered by rei
I got it wrong. I did exactly what you said. Multiplied m times g times h then multiplied by 2. Sq root that number and got 2778.926411 but it was wrong.
Answered by drwls
X = sqrt(2 M g H/k)
= 4.54 m

You forgot to divide by k, the spring constant.
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